Question: What is the value of $\dfrac{d}{dx}\left(\dfrac{1}{x}\right)$ at $x=6$ ?
Answer: The strategy We can first rewrite the fraction as a negative power of $x$. Then, the derivative can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have the derivative, we can evaluate it at $x=6$. Rewriting the fraction as a negative power $\dfrac{1}{x}=x^{-1}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(x^{-1}\right) \\\\ &=-1\cdot x^{-1-1} \gray{\text{The power rule}} \\\\ &=-x^{-2} \end{aligned}$ Evaluating the derivative So we found that $\dfrac{d}{dx}\left(\dfrac{1}{x}\right)=-x^{-2}$, which can also be written as $-\dfrac{1}{x^2}$. Now let's plug ${x=6}$ : $\begin{aligned} -\dfrac{1}{({6})^2}&=-\dfrac{1}{36} \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\left(\dfrac{1}{x}\right)$ at $x=6$ is $-\dfrac{1}{36}$.